Bunuel wrote:

Two taps can fill a cistern in 20 minutes and 30 minutes. The first tap was opened initially for x minutes after which the second tap was opened. If it took a total of 15 minutes for the tank to be filled, what is the value of x?

(A) 5.0

(B) 7.5

(C) 9.0

(D) 10.0

(E) 12.5

I backsolved here. Finding equation was taking too long.

1. Start with (C), where Tap 1 is open by itself for 9 minutes.

Tap 1 rate = \(\frac{1}{20}\)

W = rt, so \(\frac{1}{20}\) * 9 = \(\frac{9}{20}\) work is completed --->

\(\frac{11}{20}\) work

remains.

9 minutes have passed.2. Then Tap 1 and Tap 2 work together. Add rates.

\(\frac{1}{20}\) + \(\frac{1}{30}\) = \(\frac{50}{600}\) =

\(\frac{1}{12}\)W/r = t: \(\frac{11}{20}\) divided by \(\frac{1}{12}\) = \(\frac{132}{60}\) = 6.6 minutes

9 minutes elapsed + 6.6 more = 15.6 minutes. x = 9 is just a little too large. Try (B).

3. Answer (B), Tap 1 alone for 7.5 = \(\frac{15}{2}\) minutes

Work done by Tap 1: \(\frac{1}{20}\) * \(\frac{15}{2}\) = \(\frac{3}{8}\)

\(\frac{5}{8}\) of work remains. 7.5 minutes have elapsed.

4. Taps 1 and 2 work together. From above, combined rate is \(\frac{1}{12}\).

W/r = t: \(\frac{5}{8}\) divided by \(\frac{1}{12}\) =

\(\frac{15}{2}\) = 7.5 minutes.

7.5 minutes elapsed + 7.5 minutes more = 15 minutes.

Answer B.
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